Introduction

In this blog post, we will explore how to solve the LeetCode problem "2161" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed integer array nums and an integer pivot. Rearrange nums such that the following conditions are satisfied: Every element less than pivot appears before every element greater than pivot. Every element equal to pivot appears in between the elements less than and greater than pivot. The relative order of the elements less than pivot and the elements greater than pivot is maintained. More formally, consider every pi, pj where pi is the new position of the ith element and pj is the new position of the jth element. If i < j and both elements are smaller (or larger) than pivot, then pi < pj. Return nums after the rearrangement. Example 1: Input: nums = [9,12,5,10,14,3,10], pivot = 10 Output: [9,5,3,10,10,12,14] Explanation: The elements 9, 5, and 3 are less than the pivot so they are on the left side of the array. The elements 12 and 14 are greater than the pivot so they are on the right side of the array. The relative ordering of the elements less than and greater than pivot is also maintained. [9, 5, 3] and [12, 14] are the respective orderings. Example 2: Input: nums = [-3,4,3,2], pivot = 2 Output: [-3,2,4,3] Explanation: The element -3 is less than the pivot so it is on the left side of the array. The elements 4 and 3 are greater than the pivot so they are on the right side of the array. The relative ordering of the elements less than and greater than pivot is also maintained. [-3] and [4, 3] are the respective orderings. Constraints: 1 <= nums.length <= 105 -106 <= nums[i] <= 106 pivot equals to an element of nums.

Explanation

Here's the solution to rearrange the array based on the given pivot:

• Partitioning: Create three separate lists: one for elements less than the pivot, one for elements equal to the pivot, and one for elements greater than the pivot. Iterate through the input array and append each element to the appropriate list.

• Concatenation: Concatenate the three lists in the order: less than, equal to, and greater than. This ensures the desired arrangement.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def pivotArray(nums, pivot):
    less = []
    equal = []
    greater = []

    for num in nums:
        if num < pivot:
            less.append(num)
        elif num == pivot:
            equal.append(num)
        else:
            greater.append(num)

    return less + equal + greater